Let $\{x\}$ and $[x]$ denote the fractional part of $x$ and the greatest integer $\leq x$ respectively of a
real number $x$. If $\int_{0}^{n}\{x\} d x, \int_{0}^{n}[x] d x$ and
$10\left(n^{2}-n\right),(n \in N, n>1)$ are three consecutive terms of a G.P., then $\mathrm{n}$ is equal to
$\int_{0}^{n}\{x\} d x=n \int_{0}^{1}\{x\} d x=n \int_{0}^{1} x d x=\frac{n}{2}$
$\int_{0}^{n}[x] d x=\int_{0}^{n}(x-\{x\}) d x=\frac{n^{2}}{2}-\frac{n}{2}$
$\Rightarrow\left(\frac{\mathrm{n}^{2}-\mathrm{n}}{2}\right)^{2}=\frac{\mathrm{n}}{2} \cdot 10 \cdot \mathrm{n}(\mathrm{n}-1) \quad($ where $\mathrm{n}>1)$
$\Rightarrow \frac{\mathrm{n}-1}{4}=5 \Rightarrow \mathrm{n}=21$
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